Question

20 ml of 0.1N NaOH was required to neutralise a solution of a dibasic acid. Weight of the acid present in the solution is :

[Molecular weight of the acid $=$ 126]

• A. $0.63g$
• B. $63g$
• C. $0.126g$
• D. $6.3g$

Answer 1

Answer: (C)

$0.126g$

$N_1{V}_1=N_2{V}_2$

$0\cdot 1\times 20=N_2{V}_2$

$N_2{V}_2=2$

$N_2=\frac{w_2}{\frac{M\cdot wt}{2}}\times \frac{1000}{V_2\left(ml\right)}$

$N_2{V}_2=\frac{w_2}{\frac{126}{2}}\times 1000$

$2=\frac{w_2}{126}\times 2\times 1000$

$w_2=\frac{126}{1000}$

$w_2=\cdot 126g$